posted by Halle on .
I need help with the restrictions on the variables of this question:
Simplify. State any restrictions on the variables.
So my answer is: log(x+4/x-3) which is correct.
Now for the restrictions, I have:
x<-4 and x>3
However the back of the book says the answer is only x>3 and not x<-4 for the restrictions of the variables. But when I sub in any number such as -5 into the ORIGINAL log equation, i get a defined and positive log of x value. So? is mine correct or theirs? Explain fully.
Sorry the question is:
first of all
log(x^2+7x+12)/log(x^2-9) is NOT equal to
log((x+4)/(x-3)) like you say
If the back of your book has that answer, then they are definitely WRONG
would be equal to log((x+4)/(x-3))
to find the restrictions you have to consider the numerator and denominator
remember we can only take logs of positive numbers
at the top:
log(x^2+7x+12) = log[(x+3)(x+4)]
x < -4 OR x > -3
at the bottom:
log(x^2 - 9) = log[(x-3)(x+3)]
x < -3 or x > 3
but both of these conditions must be met, so
x < -4 OR x > -3 AND x < -3 or x > 3
which leaves you with x < -4 or x > 3
notice that between -4 and +3 we get a negative for either one or the other function
another way to understand the restriction would be:
graph both y = (x+3)(x+4) and
y = (x+3)(x-3)
any x where either or both of those graphs dip below the x-axis would not be allowed.
so yes, they are wrong with their restrictions.
ARGHHHH! saw your correction only after I posted.
So my answer would match your first posting, and it would clearly be a more interesting question.
= log(x+4) + log(x+3) - log(x+3) - log(x-3)
now each of those terms must be defined.
so x>-4 AND x>-3 AND x>3
which is x > 3
notice that we canceled out log(x+3)
but for all x ≤ -3 that term would be undefined. So can we really do algebra with undefined numbers??
I know this is a bit tricky, since -5 (or any x < -4) would give a positive result in the original algebraic expression, but it would not work in the individuals terms.